Geometry. Selected problems on similar triangles (from last homework).

Transcription

1 October 25, 2015 Geoetry. Selecte probles on siilar triangles (fro last hoework). Proble 1(5). Prove that altitues of any triangle are the bisectors in another triangle, whose vertices are the feet of these altitues (hint: prove that the line connecting the feet of two altitues of a triangle cuts off a triangle siilar to it). B Solution. Notice siilar right triangles,, which iplies,. Therefore,. Siilarly, fro it follows that, an fro that. Hc H Hb Ha Proble 2(2). Rectangle EFG is inscribe in triangle B such that the sie E belongs to the base B of the triangle, while points F an G belong to sies B ab, respectively. What is the largest area of rectangle EFG? Solution. Notice siilar triangles,, wherefro the vertical sie of the rectangle is, so that the area of the rectangle is, Using the geoetric-arithetic ean,. (a) (b) G G H H E F B x H (1-x) H S G+S EFB+ S = S +S = x2s +(1-x) 2 E G E B S B >= ½S B x 2+(1-x) 2 = 1-x+2x 2 = ½+2(x-½) 2) >= ½ (c) inequality,, E E where the largest value of the left sie is achieve when, an therefore. There are a nuber of other possible solutions, soe of which are shown in the figures. G F B B, E, S E= S E S +S + = su of the areas of shae triangles G EFB S >= ½S E B G F B

2 Recap: Property of the bisector. Theore (property of the bisector). The bisector of any angle of a triangle ivies the opposite sie into parts proportional to the ajacent sies,,, Proof. onsier the bisector BB. raw line parallel to BB fro the vertex, which intercepts the extension of the sie B at a point. ngles B B an B have parallel sies an therefore are congruent. Siilarly are congruent BB an B. Hence, triangle B is isosceles, an B = B. Now, applying the intercept theore to the triangles BB an, we obtain B O B. Theore (property of the external bisector). The bisector of the exterior angle of a triangle intercepts the opposite sie at a point ( in the Figure) such that the istances fro this point to the vertices of the triangle belonging to the sae line are proportional to the lateral sies of the triangle. Proof. raw line parallel to fro the vertex B, which intercepts the sie at a point B. ngles BB an B have parallel sies an therefore are congruent. Siilarly, we see that angles B an BB are congruent, an, therefore, B B. pplying the intercept theore, we obtain,. B B

3 oensurate an incoensurate segents. The Eucliean algorith. efinition. Two segents, an, are coensurate if there exists a thir segent,, such that it is containe in each of the first two segents a whole nubers of ties with no reainer.,,, s b=s s a=ns The segent is calle a coon easure of the segents an. The concept of coensurability is siilar to that of the coon ivisor for integers. It can be extene to any two quantities of the sae enoination two angles, two arcs of the sae raius, or two weights. The greatest coon easure. If a coon easure of two segents an is sub-ivie into two, three, or, generally, any nuber of equal saller segents, these saller segents are also coon easures of the segents an. In this way, an infinite set of coon easures, ecreasing in length, can be obtaine,,. Since any coon easure is less than the saller segent,, there ust be the largest aong the coon easures, which is calle the greatest coon easure. Fining the greatest coon easure (GM) is one by the etho of consecutive exhaustion calle Eucliean algorith. It is siilar to the etho of consecutive ivision use for fining the greatest coon ivisor in arithetic. The etho is base on the following theore. Theore. Two segents an are coensurate, if an only if the saller segent,, is containe in the greater one a whole nuber of ties with no reainer, or with a reainer,, which is coensurate with the saller segent,.,,.

4 The greatest coon easure of two segents is also the greatest coon easure of the saller segent an the reainer, or there is no reainer. Proof. First, consier the necessary conition. Let an be coensurate,,,, an. Let be their greatest coon easure. Then, either ( ) an segent is containe in a whole nuber of ties with no reainer, being the GM of the two segents, or,,. Then,, where, an, therefore,,, which shows that an are coensurate. The sufficiency follows fro the observations that (i) if segent is containe in a whole nuber of ties with no reainer, then the segents are coensurate, an is the greatest coon easure of the two, while (ii) if, an an are coensurate with the greatest coon easure,,, then,, an an are also coensurate with the sae GM. The Eucliean algorith. In orer to fin the GM of the two segents, an, we can procee as follows. First, using a copass exhaust the greater segent, arking on it the saller segent as any ties as possible, until the reainer is saller than the saller segent,, or there is no reainer. ccoring to rchiees exhaustion axio, these are the only two possible outcoes. Following the above theore, the proble now reuces to fining the GM of this reainer,, an the saller segent,. We now repeat the sae proceure, exhausting segent with, an again, there is either no reainer an is the GM of an, or there is a reainer. The proble is then reuce to fining the GM of a pair of even saller segents, an, an so on. If segents an are coensurate an their GM,, exists, then this process will en after soe nuber of steps, naely, on step where. Inee, all reainers in this process are ultiples of,, an is the ecreasing sequence of natural nubers, which necessarily terinates, since any non-epty set of positive integers has the sallest nuber principle of

5 the sallest integer. If the proceure never terinates, then segents have no coon easure an are incoensurate. an Exaple. The hypotenuse of an isosceles right triangle is incoensurate to its leg. Or, equivalently, the iagonal of a square is incoensurate to its sie. Proof. onsier the isosceles right triangle B shown in the Figure. Because the B hypotenuse is less than twice the leg by the E triangle inequality, the leg can only fit once in the hypotenuse, this is arke by the segent. Let the perpenicular to the hypotenuse at point intercept leg B at point E. Triangle BE is also isosceles. This is because angles BE an BE suppleent equal angles B an B to 90 egrees, an therefore are also equal. Triangle E is an isosceles right triangle, siilar to B. Its leg = - B = E = BE is a reainer of subtracting the leg B = fro the hypotenuse,, while the hypotenuse, E = B - BE = B -, is the reainer of subtracting this reainer fro the leg B = B. Hence, on the secon step of the Eucliean algorith we arrive at the sae proble as the initial one, only scale own by soe overall factor. Obviously, this process never ens, an therefore the hypotenuse an the leg B are incoensurate.

6 The Law of Lever. The Metho of the enter of Mass. rchiees Law of Lever. "Give e a place to stan on, an I will ove the earth." quote by Pappus of lexanria in Synagoge, Book VIII, c. 340 rchiees of Syracuse generally consiere the greatest atheatician of antiquity an one of the greatest of all tie. rchiees anticipate oern calculus an analysis by applying concepts of infinitesials an the etho of exhaustion to erive an rigorously prove a range of geoetrical theores, incluing the area of a circle, the surface area an volue of a sphere, an the area uner a parabola. rchiees of Syracuse Born c. 287 B Syracuse, Sicily Magna Graecia ie c. 212 B (age aroun 75), Syracuse He was also one of the first to apply atheatics to physical phenoena, founing hyrostatics an statics, incluing an explanation of the principle of the lever. He is creite with esigning innovative achines, such as his screw pup, copoun pulleys, an efensive war achines to protect his native Syracuse fro the Roan invasion. rchiees erives the Law of Lever fro several siple axios (assuptions), which suarize the everyay experience, in a anner siilar to those in Eucliean geoetry. xio 1. Equal weights at equal istances fro the fulcru balance. Equal weights at unequal istance fro the fulcru o not balance, but the weight at the greater istance will tilt its en of the lever own.

7 xio 2. If, when two weights balance, we a soething to one of the weights, they no longer balance. The sie holing the weight we increase goes own. M xio 3. If, when two weights balance, we take soething away fro one of the, they no longer balance. The sie holing the weight we i not change goes own. rchiees then proves the inverse stateents as propositions (theores). Proposition 1. Weights that balance at equal istances fro the fulcru are equal. Proposition 2. Unequal weights at equal istances fro the fulcru o not balance, but the sie holing the heavier weight goes own. Proposition 3. Unequal weights balance at unequal istances fro the fulcru, the heavier weight being at the shorter istance. Proposition 4. If two equal weights have ifferent centers of gravity then the center of gravity of the two together is the ipoint of the line segent joining their centers of gravity. 2

8 Proposition 4 is just a rephrase of the xio 1, where rchiees tacitly introuces the notion of the enter of Gravity (enter of Mass). The way to unerstan the Proposition 4 is to treat the entire weight as if it is locate at a single point, its center of gravity. In other wors, we can picture each weight (ass) as concentrate in a single point, i. e. as a Point Mass. We shall use ters weight an ass interchangeably, assuing that weight is associate with a ass in the hoogeneous gravitation fiel, an therefore is proportional to the ass. The following observation ieiately follows fro the Proposition 4. orollary. If an even nuber of equal weights have their centers of gravity situate along a straight line such that the istances between the consecutive weights are all equal, then the center of gravity of the entire syste is the ipoint of the line segents joining the centers of gravity of the two weights in the ile. M t this point rchiees proves the Law of Lever, first only for coensurate weights. Proposition 5. oensurate weights (asses) balance at istances fro the fulcru, which are inversely proportional to their agnitues,. M=5w =2w =2l 10x1/2w M=5w =5l 4x1/2w =2w 2 5

9 Proof. Let be the greatest coon easure of weights (asses) an,,,,. Let us split weight into saller pieces, each of weight, an weight into saller pieces of weight. Let us now split the segent connecting an into congruent saller segents, an also ark such segents on the opposite sie of weight an such segents on the opposite sie of weight. Let us now place all saller weights at the centers of these segents as shown in the Figure. learly, since each of the initial weights was split into an even nuber of equal pieces, which were place syetrically aroun its initial position, the resultant syste of saller weights has the sae center of gravity as the original weight. On the other han, the obtaine syste of weights has the center of gravity in the ile, at a istance of segents fro the position of weight an segents fro the position of weight, as illustrate in the Figure. Therefore,, which proves the Law of Lever for the coensurate weights. The theore for the incoensurate weights is then proven by reucing to contraiction. Theore (Law of Lever). Incoensurate weights (asses) balance at istances fro the fulcru, which are inversely proportional to their agnitues, Proof. Let weights an be place at istances an fro the fulcru, respectively, such that the Law of Lever is satisfie,. ssue that the weights nevertheless o not balance, for exaple, goes own. Reove a sall aount fro weight, turning it into weight, such that it still goes own, but is now coensurate with. Now an are coensurate, an, which eans that shoul rise. This contraicts our assuption, so an ust balance. Note that in the above rchiees iplies a non-trivial fact that a coensurate weight can be foun that iffers fro the given incoensurate weight by an arbitrarily sall aount. This eans that for any irrational nuber there exists a rational nuber, which iffers fro it as little as we want, i. e. that rational nubers are ense.

10 Metho of the enter of Mass (Mass Points). efinition. For two point asses, an B at points an B, the center of ass lies at a point on the straight line segent B, such that,. When fining the center of ass in a syste of point asses, one can replace any pair of asses, an B, with a single point ass having the total ass + B, place at the center of ass of the pair. The following iportant properties of the enter of Mass follow ieiately. 1. Every syste of finite nuber of point asses has unique center of ass (OM). 2. For two point asses, the OM belongs to the segent connecting these points; its position is eterine by the rchiees lever rule: the point s ass ties the istance fro it to the OM is the sae for both points. 3. The position of the syste s center of ass oes not change if we ove any subset of point asses in the syste to the center of ass of this subset. In other wors, we can replace any nuber of point asses with a single point ass, whose ass equals the su of all these asses an which is positione at their OM. eva s Theore: Point Masses. We select asses,, B, an such that the corresponing centers of ass for each pair are at points, B an, respectively. Then,. B + B B O B B + +